A secant line intersects the curve $y=\tan(x)$ at two points with $x$ -coordinates $4$ and $4+h$. What is the slope of the secant line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\tan(4+h)-\tan(4)}{4}$ (Choice B) B $\dfrac{\tan(4+h)-\tan(4)}{h}$ (Choice C) C $\dfrac{\tan(4+h)-\tan(4)}{4+h}$ (Choice D) D $\dfrac{\tan(h)-\tan(4)}{h}$
When $x=4$, then $y=\tan(4)$. When $x=4+h$, then $y=\tan(4+h)$. The secant line goes through the points $(4, \tan(4))$ and $(4+h, \tan(4+h))$. This is enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{\tan(4+h)-\tan(4)}{4+h-4} \\\\ &=\dfrac{\tan(4+h)-\tan(4)}{h} \end{aligned}$ The slope of the secant line: $\dfrac{\tan(4+h)-\tan(4)}{h}$